If `vecA=2veci+3vecj+4veck and vecB=4veci+3vecj+2veck, find vecAxxvecB`.

Explanation

To find the cross product of vectors \(\vec{A}\) and \(\vec{B}\), we can follow these steps:

1. **Write down the vectors**:
\[ \vec{A} = 2\hat{i} + 3\hat{j} + 4\hat{k} \]
\[ \vec{B} = 4\hat{i} + 3\hat{j} + 2\hat{k} \]

2. **Set up the determinant for the cross product**:
The cross product \(\vec{A} \times \vec{B}\) can be calculated using the determinant of a matrix formed by the unit vectors \(\hat{i}, \hat{j}, \hat{k}\) and the components of \(\vec{A}\) and \(\vec{B}\):
\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 4 & 3 & 2 \end{vmatrix} \]

3. **Calculate the determinant**:
Using the determinant formula:
\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 3 & 4 \\ 3 & 2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 4 \\ 4 & 2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 4 & 3 \end{vmatrix} \]

Now, we calculate each of the 2x2 determinants:
- For \(\hat{i}\):
\[ \begin{vmatrix} 3 & 4 \\ 3 & 2 \end{vmatrix} = (3)(2) - (4)(3) = 6 - 12 = -6 \]

- For \(\hat{j}\):
\[ \begin{vmatrix} 2 & 4 \\ 4 & 2 \end{vmatrix} = (2)(2) - (4)(4) = 4 - 16 = -12 \]

- For \(\hat{k}\):
\[ \begin{vmatrix} 2 & 3 \\ 4 & 3 \end{vmatrix} = (2)(3) - (3)(4) = 6 - 12 = -6 \]

4. **Substituting back into the equation**:
Now substituting the values back into the equation:
\[ \vec{A} \times \vec{B} = -6\hat{i} - (-12)\hat{j} - 6\hat{k} \]
\[ = -6\hat{i} + 12\hat{j} - 6\hat{k} \]

5. **Final result**:
Therefore, the cross product \(\vec{A} \times \vec{B}\) is:
\[ \vec{A} \times \vec{B} = -6\hat{i} + 12\hat{j} - 6\hat{k} \]